RD Sharma Class 12 Exercise 3.9 Inverse Trigonometric Functions Solutions Maths - Download PDF Free Online

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RD Sharma Class 12 Exercise 3.9 Inverse Trigonometric Functions Solutions Maths - Download PDF Free Online

RD Sharma Class 12 Exercise 3.9 Inverse Trigonometric Functions Solutions Maths - Download PDF Free Online

Kuldeep MauryaUpdated on 21 Jan 2022, 12:37 PM IST

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RD Sharma Class 12 Solutions Chapter 3 Inverse Trigonometric Functions - Other Exercise

Inverse Trigonometric Functions Excercise: 3.9

Inverse Trigonometric Function exercise 3.9 question 1 (i)

Answer:
$\frac{24}{25}$
Hint:
Take the negative sign out of the sine inverse bracket and after that cosine of a negative angle will be equal to cosine of the same positive angle.
This way you get rid of the negative sign
Concept:
Inverse Trigonometry
Solution:
$cos\, cos((\frac{-7}{25}))=cos\, cos(-(\frac{7}{25}))$ $(-x)=-(x)$
$=cos\, cos((\frac{7}{25}))$ $[cos\, cos(-\theta )=cos\, \theta ]$
$Let\: \: \: sin^{-1}(\frac{7}{25})=x$.......(1)
$sin\, x=\frac{7}{25}$
$cos\, cos\, x=\sqrt{1-x}$ $[x+x=1]$
$=\sqrt{1-(\frac{7}{25})^{2}}$
$=\sqrt{1-\frac{49}{625}}$
$=\sqrt{\frac{625-49}{625}}$
$=\sqrt{\frac{576}{625}}=\frac{24}{25}$
$\therefore x=cos^{-1}(\frac{24}{25})$.......(2)
From (1) and (2)
$sin^{-1}(\frac{7}{25})=cos^{-1}(\frac{24}{25})$
$\therefore cos(sin^{-1}(\frac{7}{25}))=cos(cos^{-1}(\frac{24}{25}))$
$\therefore cos(sin^{-1}(\frac{7}{25}))=\frac{24}{25}$
Note: Try to convert the inner inverse term to the inverse of outer trigo ratio or the trigo ratio that can be related directly to the outer trigo ratio using identities.

Inverse Trigonometric Function exercise 3.9 question 1 (ii)

Answer:
$\frac{13}{5}$
Hint:
Convert the cot^-1 term to a tan^-1 term so that we can us the following result to solve the sum
$sec^{2}(x)=1+tan^{2}(x)$
$\therefore sec\, (x)=\sqrt{1+tan^{2}(x)}$ ...using this we can simplify the sum
Concept:
Inverse Trigonometry
Solution:
$sec[cot^{-1}(\frac{-5}{12})]=sec[tan^{-1}(-\frac{12}{5})]$ $[cot^{-1}(x)=tan^{-1}(\frac{1}{x})]$
$=sec[-tan^{-1}(\frac{12}{5})]$ $[tan^{-1}(-x)=-tan^{-1}(x)]$
$=sec[tan^{-1}(\frac{12}{5})]$ $[sec(-\theta )=sec\, \theta ]$
$=\sqrt{1+tan^{2}(tan^{-1}(\frac{12}{5}))}$ $[sec^{2}\theta =1+tan^{2}\theta ]$
$=\sqrt{1+(\frac{12}{5})^{2}}\; \; \; =>\sqrt{1+\frac{144}{25}}\; \; \; =\sqrt{\frac{25+144}{25}}$
$=\sqrt{\frac{169}{25}}$
$=\frac{13}{5}$
Note: Try to convert the inner inverse term to the inverse of outer trigo ratio or the trigo ratio that can be related directly to the outer ratio using identities

Inverse Trigonometric Function exercise 3.9 question 1 (iii)

Answer:
$\frac{-5}{12}$
Hint:
Eliminate the negative sign by using formula sec^-1(-x)=$\pi$-sec^-1(x) and then convert the sec^-1 term into a cot^-1 term to simplify the sum
Concept:
Inverse Trigonometric
Solution:
$cot(sec^{-1}(\frac{-13}{5}))=cot(\pi -sec^{-1}(\frac{13}{5}))$ $[sec^{-1}(-x)=\pi -sec^{-1}(x)]$
$=-cot(sec^{-1}(\frac{13}{5}))$ $[cot(\pi -\theta) =-cot\, \theta ]$
Let
$sec^{-1}(\frac{13}{5})=x$ ..............(1)
$sec\, x=\frac{13}{5}$
$\therefore tan\, x=\sqrt{sec^{2}x-1}$
$=\sqrt{(\frac{13}{5})^{2}-1}\: \: \: =\frac{12}{5}$
$cot\, x=\frac{5}{12}\Rightarrow x=cot^{-1}(\frac{5}{12})$ ............(2)
$sec^{-1}(\frac{13}{5})=cot^{-1}(\frac{5}{12})$
From (1) and (2)
$=cot(sec^{-1}(\frac{-13}{5}))$
$=-cot(sec^{-1}(\frac{13}{5}))$
$=-cot(cot(\frac{5}{12}))$
$=\frac{-5}{12}$
Note: Try to convert the inner inverse term to the inverse of outer trigo ratio or the trigo ratio that can be related directly to the outer ratio using identities

Inverse Trigonometric Function exercise 3.9 question 2 (i)

Answer:
$\frac{-24}{7}$
Hint:
Get rid of the negative sign by using the formula
$cos^{-1}(-x)=\pi -cos^{-1}(x)$
Convert the cos^-1 term to a sec^-1 term.
Now use the formula
$sec^{2}(x)=1+tan^{2}(x)$
$tan(x)=\sqrt{sec^{2}(x)-1}$
Concept:
Inverse Trigonometry
Solution:
$tan(cos^{-1}(\frac{-7}{25}))=tan(\pi -cos^{-1}(\frac{7}{25}))$ $[cos^{-1}(-x)=\pi -cos^{-1}(x)]$
$=-tan(\pi -cos^{-1}(\frac{7}{25}))$ $[tan(\pi -\theta )=-tan\theta ]$
$=-tan(sec^{-1}(\frac{25}{7}))$ $[cos^{-1}(x)=sec^{-1}(\frac{1}{x})]$
$=-\sqrt{sec^{2}(sec^{-1}(\frac{25}{7}))-1}$ $[tan^{2}\theta =sec^{2}\theta -1]$
$=-\sqrt{\frac{625-49}{49}}$ $=-\frac{24}{7}$
Note: Try to convert the inner inverse term to the inverse of outer trigo ratio or the trigo ratio that can be related directly to the outer ratio using identities

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Inverse Trigonometric Function exercise 3.9 question 2 (ii)

Answer:
$\frac{13}{5}$
Hint:
Get rid of the negative sign by using the formula
$cos^{-1}(-x)=\pi -cot^{-1}(x)$
Convert the cos^-1term to a sec^-1 term.
Now use the formula
$cos\, ec^{2}(x)=1+cot^{2}(x)$
$cos\, ec(x)=\sqrt{cot^{2}(x)+1}$
Concept:
Inverse Trigonometry
Solution:
Let
$cot^{-1}(\frac{-12}{5})=x$
$cot\, x=\frac{-12}{5}$
$cos\, ec^{2}(x)=1+cot^{2}(x)$
$cos\, ec(x)=\sqrt{cot^{2}(x)+1}$
$=\sqrt{1+(\frac{12}{5})^{2}}$
$=\sqrt{1+(\frac{144}{25})}$
$=\sqrt{\frac{169}{25}}\; \; \; \; \;= \frac{13}{5}$
$cos\, ec(x)= \frac{13}{5}$
$x=cot^{-1} (\frac{-12}{5})$
$cos\, ec(cot^{-1} (\frac{-12}{5}))=\frac{13}{5}$
Note: Try to convert the inner inverse term to the inverse of outer trigo ratio or the trigo ratio that can be related directly to the outer ratio using identities

Inverse Trigonometric Function exercise 3.9 question 2 (iii)

Answer:
$\frac{4}{5}$
Hint:
Get rid of the negative sign by using the property
$tan^{-1}(-x)=-tan^{-1}(x)$
Now use this property to simplify
$cos(-x)=cos(x)$
Now find the sec^-1 equivalent of the given tan^-1 term
Now use
$cos(x)=\frac{1}{sec (x)}$$cos(x)=\frac{1}{sec (x)}.\: \: to \; solve\: the\: sum$
Concept:
Inverse Trigonometry
Solution:
Let
$tan^{-1}(\frac{-3}{4})=x$
$tan\, x=\frac{-3}{4}$
$sec^{2}x=1+tan^{2}x$
$sec\, x=\sqrt{1+tan^{2}x}$
$=\sqrt{1+(\frac{-3}{4})^{2}}$
$=\sqrt{1+\frac{9}{16}}\; \; \: =\sqrt{\frac{25}{16}}$
$\therefore sec(x)=\frac{5}{4}$
$\therefore cos(x)=\frac{1}{sec(x)}\; \; \; =\frac{4}{5}$
$But \; \; x=tan^{-1}(\frac{-3}{4})$
$cos(tan^{-1}(\frac{-3}{4}))=\frac{4}{5}$
Note: Try to convert the inner inverse term to the inverse of outer trigo ratio or the trigo ratio that can be related directly to the outer ratio using identities

Inverse Trigonometric Function exercise 3.9 question 3

Answer:
$\frac{-56}{65}$
Hint:
Get rid of the negative sign by using the property
$cot^{-1}(-x)=\pi -cot^{-1}(x)$
$cos^{-1}(-x)=\pi -cost^{-1}(x)$
Use formula
$sin(A+B)=sinAcosB+cosAsinB$
Concept:
Inverse Trigonometry
Solution:
Let
$cos^{-1}\frac{3}{5}=\theta _{1}$ ....(1)
$cos \: \theta _{1}=\frac{3}{5}$
$\theta _{1}=sin^{-1}(\frac{4}{5})$ ....(2)

Let
$cot^{-1}(\frac{5}{12})=\theta _{2}$ ....(3)
$cot\, \theta _{2}=\frac{5}{12}$
$\theta _{2}=cos^{-1}(\frac{5}{13})\; \; =sin^{-1}(\frac{12}{13})$ ....(4)

$sin\, sin((\frac{-3}{5})+cot\, cot(\frac{-5}{12}))$
$=sin\, sin(\pi -(\frac{3}{5})+\pi -(\frac{5}{12}))$
$=sin\, (2\pi -(cos^{-1}(\frac{3}{5})+cot^{-1}(\frac{5}{12})))$
$=-sin\, (cos^{-1}(\frac{3}{5})+cot^{-1}(\frac{5}{12}))$
$=-\left[\sin \left(\cos ^{-1}\left(\frac{3}{5}\right)\right) \cdot \cos \left(\cot ^{-1}\left(\frac{5}{12}\right)\right)+\cos \left(\cos ^{-1}\left(\frac{3}{5}\right)\right) \cdot \sin \left(\cot ^{-1}\left(\frac{5}{12}\right)\right)\right]$
$=-\left[\sin \left(\sin ^{-1}\left(\frac{4}{5}\right)\right) \cdot \cos \left(\cos ^{-1}\left(\frac{5}{13}\right)\right)+\cos \left(\cos ^{-1}\left(\frac{3}{5}\right)\right) \cdot \sin \left(\sin ^{-1}\left(\frac{12}{13}\right)\right)\right]$
From (1), (2), (3) and (4)
$=-[\frac{4}{5}\times \frac{5}{13}+\frac{3}{5}\times\frac{12}{13} ]$
$=[\frac{-56}{65} ]$

Note: Try to convert the inner inverse term to the inverse of outer trigo ratio or the trigo ratio that can be related directly to the outer ratio using identities

tion of all trigonometric functions and their simplification.

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